Determining if a precipitate will form from a double replacement reaction.

 

We will use the example of combining solutions of K2CrO4 and NaOH as an example. The first thing that needs to be done is write out the possible products of these two reactants. For help in writing the correct formulas for product on double replacement reactions click here.

 

NaOH    +     K2CrO4     ®      KOH    +    Na2CrO4

 

We can assume that the reactants are both in an aqueous solution because it said so in the question above. The way this is written in an equation is by placing (aq) to the right of the formula. “aq” is short for aqueous or a essentially the compound in question dissolved in a solvent making a solution.

 

NaOH(aq)   +     K2CrO4(aq)   ®      KOH    +    Na2CrO4

 

The next step is to determine if one of the products will form a precipitate. This means that it is insoluble in water and will form a solid if in water. The way to determine this is by using the solubility rules.

 

  1. Most nitrate (NO3-) salts are soluble.
  2. Most salts of Na+, K+, and NH4+ are soluble.
  3. Most chloride salts are soluble. Notable exceptions are AgCl, PbCl2, and Hg2Cl2
  4. Most sulfate salts are soluble. Notable exceptions are BaSO4, PbSO4, and CaSO4.
  5. Most hydroxide compounds are only slightly soluble.* The important exceptions are NaOH and KOH. Ba(OH)2 and Ca(OH)2 are only moderately soluble.
  6. Most sulfide (S-), carbonate (CO32-), and phosphate (PO42-) salts are only slightly soluble. *

 

*The terms insoluble and slightly soluble really mean the same thing: such a tiny amount dissolves that it is not possible to detect it with the naked eye.

 
 
 

 

 

 

 

 

 

 

 

 

 

 

 

 


Look at the cation and anions independently and run each through the rules. Lets run the KOH through the rules first. “K” falls under the 2nd rule which indicates that its is soluble without exception. Therefore we determine that KOH is soluble. If you happened to miss this, rule 5 states that compounds with “OH” are insoluble except when combined with “K”. This restates that KOH is soluble.

 

We cannot forget the second product “Na2CrO4”. Looking at the cation “Na” we find that it is soluble without exception in rule number 2. Therefore both KOH and Na2CrO4 will both remain soluble in water. The final equation will than be written as follows:

 

2 NaOH(aq)   +     K2CrO4(aq)   ®     2 KOH (aq)   +    Na2CrO4 (aq)

 

Notice that the equation is also balanced.